Exercise 8.2 - Chapter 8 Binomial Theorem class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 1:

Find the coefficient of x⁵ in (x + 3)⁸

Answer:

It is known that (r + 1)^th term, (Tr₊₁), in the binomial expansion of (a + b)n is given by  .

Assuming that x⁵ occurs in the (r + 1)^th term of the expansion (x + 3)⁸, we obtain

Comparing the indices of x in x⁵ and in Tr ₊₁, we obtain

r = 3

Thus, the coefficient of x⁵ is

Question 2:

Find the coefficient of a⁵b⁷ in (a – 2b)¹²

Answer:

It is known that (r + 1)^th term, (Tr₊₁), in the binomial expansion of (a + b)n is given by  .

Assuming that a⁵b⁷ occurs in the (r + 1)^th term of the expansion (a – 2b)¹², we obtain

Comparing the indices of a and b in a⁵ b⁷ and in Tr ₊₁, we obtain

r = 7

Thus, the coefficient of a⁵b⁷ is 

Question 3:

Write the general term in the expansion of (x² – y)⁶

Answer:

It is known that the general term Tr₊₁ {which is the (r + 1)^th term} in the binomial expansion of (a + b)n is given by  .

Thus, the general term in the expansion of (x² – y⁶) is

Question 4:

Write the general term in the expansion of (x² – yx)¹², x ≠ 0

Answer:

It is known that the general term Tr₊₁ {which is the (r + 1)^th term} in the binomial expansion of (a + b)n is given by  .

Thus, the general term in the expansion of(x² – yx)¹² is

Question 5:

Find the 4^th term in the expansion of (x – 2y)¹² .

Answer:

It is known that (r + 1)^th term, (Tr₊₁), in the binomial expansion of (a + b)n is given by  .

Thus, the 4^th term in the expansion of (x – 2y)¹² is

Question 6:

Find the 13^th term in the expansion of .

Answer:

It is known that (r + 1)^th term, (Tr₊₁), in the binomial expansion of (a + b)n is given by  .

Thus, 13^th term in the expansion of  is

Question 7:

Find the middle terms in the expansions of 

Answer:

It is known that in the expansion of (a + b)n, if n is odd, then there are two middle terms, namely,  term and  term.

Therefore, the middle terms in the expansion of  are  term and  term

Thus, the middle terms in the expansion of  are  .

Question 8:

Find the middle terms in the expansions of 

Answer:

It is known that in the expansion (a + b)n, if n is even, then the middle term is  term.

Therefore, the middle term in the expansion of  is  term

Thus, the middle term in the expansion of  is 61236 x⁵y⁵.

Question 9:

In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.

Answer:

It is known that (r + 1)^th term, (Tr₊₁), in the binomial expansion of (a + b)n is given by  .

Assuming that am occurs in the (r + 1)^th term of the expansion (1 + a)m ⁺ n, we obtain

Comparing the indices of a in am and in Tr _{+ 1}, we obtain

r = m

Therefore, the coefficient of am is

Assuming that an occurs in the (k + 1)^th term of the expansion (1 + a)m⁺n, we obtain

Comparing the indices of a in an and in Tk _{+ 1}, we obtain

k = n

Therefore, the coefficient of an is

Thus, from (1) and (2), it can be observed that the coefficients of am and an in the expansion of (1 + a)m ⁺ n are equal.

Question 10:

The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of

(x + 1)n are in the ratio 1:3:5. Find n and r.

Answer:

It is known that (k + 1)^th term, (Tk₊₁), in the binomial expansion of (a + b)n is given by  .

Therefore, (r – 1)^th term in the expansion of (x + 1)n is 

r ^th term in the expansion of (x + 1)n is 

(r + 1)^th term in the expansion of (x + 1)n is 

Therefore, the coefficients of the (r – 1)^th, r^th, and (r + 1)^th terms in the expansion of (x + 1)n are   respectively. Since these coefficients are in the ratio 1:3:5, we obtain

Multiplying (1) by 3 and subtracting it from (2), we obtain

4r – 12 = 0

⇒ r = 3

Putting the value of r in (1), we obtain

n – 12 + 5 = 0

⇒ n = 7

Thus, n = 7 and r = 3

Question 11:

Prove that the coefficient of xn in the expansion of (1 + x)²n is twice the coefficient of xn in the expansion of (1 + x)²n^–1 .

Answer:

It is known that (r + 1)^th term, (Tr₊₁), in the binomial expansion of (a + b)n is given by  .

Assuming that xn occurs in the (r + 1)^th term of the expansion of (1 + x)²n, we obtain

Comparing the indices of x in xn and in Tr _{+ 1}, we obtain

r = n

Therefore, the coefficient of xn in the expansion of (1 + x)²n is

Assuming that xn occurs in the (k +1)^th term of the expansion (1 + x)²n ^{– 1}, we obtain

Comparing the indices of x in xn and Tk _{+ 1}, we obtain

k = n

Therefore, the coefficient of xn in the expansion of (1 + x)²n ^–1 is

From (1) and (2), it is observed that

Therefore, the coefficient of xn in the expansion of (1 + x)²n is twice the coefficient of xn in the expansion of (1 + x)²n^–1.

Hence, proved.

Question 12:

Find a positive value of m for which the coefficient of x² in the expansion

(1 + x)m is 6.

Answer:

It is known that (r + 1)^th term, (Tr₊₁), in the binomial expansion of (a + b)n is given by  .

Assuming that x² occurs in the (r + 1)^th term of the expansion (1 +x)m, we obtain

Comparing the indices of x in x² and in Tr _{+ 1}, we obtain

r = 2

Therefore, the coefficient of x² is .

It is given that the coefficient of x² in the expansion (1 + x)m is 6.

Thus, the positive value of m, for which the coefficient of x² in the expansion

(1 + x)m is 6, is 4.